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(C) The reaction of guaiacol ($2$-methoxyphenol) with $HBr$ involves the cleavage of the ether bond. In this reaction,the oxygen atom of the methoxy g

Vedclass · Accepted Answer

Catechol (benzene$-1,2-$diol)

Vedclass · Answer

(C) The reaction of guaiacol ($2$-methoxyphenol) with $HBr$ involves the cleavage of the ether bond. In this reaction,the oxygen atom of the methoxy group gets protonated by $HBr$ to form an oxonium ion. The $Br^-$ ion then attacks the methyl group $(CH_3)$ because the $C-O$ bond between the benzene ring and the oxygen atom has partial double bond character due to resonance,making it resistant to nucleophilic attack. Thus,the $C-O$ bond between the methyl group and the oxygen atom breaks,resulting in the formation of catechol (benzene$-1,2-$diol) and methyl bromide $(CH_3Br)$.

Column-$I$ (Reactants)	Column-$II$ (Products)
$(A) \ CH_3CH_2CH_2-O-CH_3 + HI \rightarrow$	$(i) \ (CH_3)_3C-I + CH_3CH_2OH$
$(B) \ C_6H_5OCH_2CH_3 + HBr \rightarrow$	$(ii) \ CH_3CH_2CH_2OH + CH_3I$
$(C) \ (CH_3)_3C-OCH_2CH_3 + HI \rightarrow$	$(iii) \ CH_3CH_2I + C_6H_5OH$
$(D) \ C_6H_5OCH_2CH_3 + \text{conc. } H_2SO_4 + \text{conc. } HNO_3 \rightarrow$	$(iv) \ p-NO_2-C_6H_4-OCH_2CH_3$

Product $(A)$ of this reaction is:

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